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Q108P

Expert-verifiedFound in: Page 1115

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**White light (consisting of wavelengths from ${\mathbf{400}}{\mathbf{}}{\mathit{n}}{\mathit{m}}$ to ${\mathbf{700}}{\mathbf{}}{\mathit{n}}{\mathit{m}}$ ) is normally incident on a grating. Show that, no matter what the value of the grating spacing ${\mathit{d}}$ , the second order and third order overlap.**

It is proved no matter what the value of the grating spacing ** $d$**, the second order and third order overlap.

White light has wavelengths between

${\lambda}_{1}=400nm$

${\lambda}_{2}=700nm$

**An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating. **

The angular distance of the ${m}_{th}$ order diffraction maxima is given by

${\mathit{d}}{\mathit{s}}{\mathit{i}}{\mathit{n}}{\mathit{\theta}}{\mathbf{=}}{\mathit{m}}{\mathit{\lambda}}$ …(i)

Here, $\lambda $is the wavelength of the incident light and $d$ is the separation between two lines in the grating.

From equation (i) the angular distance of second order maxima from ${\lambda}_{2}$ is

$\begin{array}{l}dsin{\theta}_{2}=2x70\mathrm{mmm}\\ sin{\theta}_{2}=\frac{1400\mathrm{mm}}{d}\end{array}$

From equation (i) the angular distance of third order maxima from ${\lambda}_{1}$ is

$\begin{array}{l}dsin{\theta}_{3}=3x400\mathrm{mm}\\ sin{\theta}_{3}=\frac{1200\mathrm{mm}}{d}\end{array}$

Thus

$\mathrm{sin}{\theta}_{2}>\mathrm{sin}{\theta}_{3}$

Hence the angular distance of a part of the second order from white light is greater than the angular distance of a part of the third order. Thus the second and third order overlap.

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